J. Korean Math. Soc. 49 (2012), No. 4, pp. 733–743

http://dx.doi.org/10.4134/JKMS.2012.49.4.733

HOLONOMY DISPLACEMENTS IN THE HOPF BUNDLES

OVER CHn AND THE COMPLEX HEISENBERG GROUPS

Younggi Choi and Kyung Bai Lee

Abstract. For the “Hopf bundle” S1 → S2n,1 → CHn, horizontal liftsof simple closed curves are studied. Let γ be a piecewise smooth, simple

closed curve on a complete totally geodesic surface S in the base space.Then the holonomy displacement along γ is given by

V (γ) = eλA(γ)i,

where A(γ) is the area of the region on the surface S surrounded by γ;

λ = 1/2 or 0 depending on whether S is a complex submanifold or not.We also carry out a similar investigation for the complex Heisenberg

group R → H2n+1 → Cn.

1. Introduction

Consider the Hopf fibration S1 → S3 → S2. Let γ be a simple closed curveon S2. Pick a point in S3 over γ(0), and take the unique horizontal lift γ of γ.Since γ(1) = γ(0), γ(1) lies in the same fiber as γ(0) does. We are interestedin understanding the difference between γ(0) and γ(1). The following equalitywas already known [2]:

V (γ) = e12A(γ)i,

where V (γ) is the holonomy displacement along γ, and A(γ) is the area of theregion surrounded by γ.

In this paper, we shall generalize this fact to (higher dimensional) pseudo-spheres and the complex Heisenberg group. First we look at the fibration ofthe pseudo-sphere S2n,1

S1 → S2n,1 → CHn

Received January 31, 2011.2010 Mathematics Subject Classification. 53C29, 53C30, 32Q45.Key words and phrases. holonomy displacement, complex hyperbolic space, complex

Heisenberg group.The first author was supported by Basic Science Research Program through the National

Research Foundation of Korea funded by the Ministry of Education, Science and Technology(2010-0016673).

c⃝2012 The Korean Mathematical Society

733

734 YOUNGGI CHOI AND KYUNG BAI LEE

a principal S1-bundle over the complex hyperbolic space CHn. Let S be acomplete totally geodesic surface in the base space CHn, and ξS be the pull-back bundle over S. Let γ be a piecewise smooth, simple closed curve on Sparametrized by 0 ≤ t ≤ 1, and γ its horizontal lift. The pullback over thecurve γ is called a Hopf torus (so γ is a curve on the Hopf torus). Then

γ(1) = e12A(γ)i · γ(0) or γ(0),

depending on whether S is a complex submanifold or not, where A(γ) is thearea of the region on the surface S surrounded by γ (See Theorem 3.3).

We also carry out a similar investigation for the complex Heisenberg group.Let 1 → R → H2n+1 → Cn → 1 be the central short exact sequence of thecomplex Heisenberg group. Let S be a complete totally geodesic plane in Cn,and ξS be the pullback bundle over S. Let γ be a piecewise smooth, simpleclosed curve on S. Then

V (γ) = e(ξS) ·A(γ),

where A(γ) is the area of the region on the surface S surrounded by γ, andthe number e(ξS) is determined by the equality [v,w] = e(ξS)e2n+1 for anorthonormal basis v,w for the tangent space of S (See Theorem 4.1).

2. Preliminaries

The proof of the statement in the introduction (in the case of the Hopffibration S1 → S3 → S2) uses the Gauss-Bonnet theorem. For S1 → S2,1 →CH1, such is not available because the base space is not compact. Therefore, wecannot apply the arguments in [2] directly, and need to develop a new methodof proof. It turns out that S2,1 is the building blocks for higher dimensionalcases.

Let F → Ep→ B be a principal F -bundle (F = R1 or S1) of Riemannian

manifolds, with B a 2-dimensional complete manifold and p a Riemanniansubmersion. For a simple closed cure γ(t), 0 ≤ t ≤ 1 on B, the holonomydisplacement V (γ) along γ is defined as follows: Let γ(t) be the horizontal liftof γ. Then

γ(1) = V (γ) · γ(0)

for some V (γ) ∈ F . We shall establish a technical lemma which will be usedlater.

Lemma 2.1. Suppose V (γ) = λ(γ)A(γ) (F = R1), or eλ(γ)A(γ)i (F = S1) fora constant λ(γ), where A(γ) is the area of the region on B surrounded by apiecewise smooth simple closed curve γ. If λ(γ) is constant for all γ’s whichare the boundaries of rectangular regions, then it is constant for every piecewisesmooth simple closed curve γ.

HOLONOMY DISPLACEMENTS IN HOPF BUNDLES 735

Proof. Let us assume that F = R. The case of F = S1 will be similar. Letγ0 be a curve on B. Since the region surrounded by γ0 is compact, we mayassume that this region is contained completely in one local patch. Let

φ : R2 → U ⊂ B

be a local chart, and p−1(U) ≈ U × F . For notational simplicity, we shallidentify R2 with U (and suppress φ). Let Ω(U, γ0(0)) and Ω(R, 0) be thespace of paths emanating from γ0(0) and 0 ∈ R, respectively. For each γ ∈Ω(U, γ0(0)), let ωγ be the unique curve in R so that η(t) = γ(t) · ωγ(t) is thehorizontal lift of γ. This defines a map

Z : Ω(U) −→ Ω(R)

by Z(γ)(t) = ωγ(t). We use the sup metrics ρ on both Ω(U) and Ω(R). Thatis,

ρ(γ1, γ2) = supt∈[0,1]

d(γ1(t), γ2(t)),

where d is the distance function on U . A similar definition holds for Ω(R). Wewish to show that Z is continuous at γ0. Let ϵ > 0 be given. By the continuityof the connection, for each t ∈ [0, 1], there is an open neighborhood W of γ(t)such that any piecewise smooth curve in W has a horizontal lift which liesin W × (−ϵ/2, ϵ/2). Since γ0(I) is compact, we can find δ > 0 such that ifρ(γ0, γ) < δ, then ρ(ωγ0 , ωγ) < ϵ. This proves that Z is continuous.

Any piecewise smooth simple closed curve can be approximated by a se-quence of piecewise linear curves which are sums of boundaries of rectangularregions. Since λ(γ) is constant for rectangular regions, the same is true for anypiecewise smooth simple closed curve.

Next, we need to know all complete totally geodesic submanifolds of the basespace of the principal bundle S1 → S2n,1 → CHn. Since S2n,1 is a symmetricspace, the following gives a complete answer.

Proposition 2.2 ([1, XI Theorem 4.3]). Let (G,H, σ) be a symmetric spaceand g = h + m the canonical decomposition. Then there is a natural one-to-one correspondence between the set of linear subspaces m′ of m such that[[m′,m′],m′] ⊂ m′ and the set of complete totally geodesic submanifolds M ′

through the origin 0 of the affine symmetric space M = G/H, the correspon-dence being given by m′ = T0(M

′).

3. The bundle S1 → S2n,1 → CHn

We shall study the bundle

U(1) −→ U(1, n)/U(n)p−→ U(1, n)/(U(1)× U(n)).

Note that U(1, n)/U(n) ∼= S2n,1, and U(1, n)/(U(1) × U(n)) ∼= CHn, whereS2n,1 = H1,2n = (z0, . . . , zn) ∈ Cn : −|z0|2 +

∑ni=1 |zi|2 = −1. For more

736 YOUNGGI CHOI AND KYUNG BAI LEE

information on S2n,1, see [3]. We first consider the case when n = 1. Ratherthan using U(1)→ U(1, 1)/U(1)→ U(1, 1)/(U(1)× U(1)), we shall use

U(1)→ SU(1, 1)→ SU(1, 1)/U(1).

Here SU(1, 1) = A ∈ GL(2,C) : AJA∗ = J and det(A) = 1 where J =[−1 00 1

].

From now on, we shall use the convention of gl(n,C) ⊂ gl(2n,R) by

[z11 z12z21 z22

]−→

x11 −y11 x12 −y12y11 x11 y12 x12

x21 −y21 x22 −y22y21 x21 y22 x22

.

The group SU(1, 1) has the following natural representation into GL(4,R):

w =

w1 w2 w3 w4

−w2 w1 −w4 w3

w3 −w4 w1 −w2

w4 w3 w2 w1

with the condition w2

1 + w22 − w2

3 − w24 = 1. In fact, the map

w1 + w2i+ w3j + w4k 7−→ w

is a monomorphism from the unit quaternions into GL(4,R). Therefore,SU(1, 1) ∼= S2,1.

The circle group

S1 =

[eiz 00 e−iz

]: 0 ≤ z ≤ 2π

is a subgroup of SU(1, 1), and acts on SU(1, 1) as right translations, freelywith quotient CH1, the complex hyperbolic line, giving rise to the fibration

S1 −→ SU(1, 1) −→ CH1.

In order to understand the projection map better, let w be the “i-conjugate”of w (replace w2 by −w2). That is,

w =

w1 −w2 w3 w4

w2 w1 −w4 w3

w3 −w4 w1 w2

w4 w3 −w2 w1

.

Then,

ww =

w2

1 + w22 + w2

3 + w24 0 2(w1w3 − w2w4) 2(w2w3 + w1w4)

0 w21 + w2

2 + w23 + w2

4 −2(w2w3 + w1w4) 2(w1w3 − w2w4)2(w1w3 − w2w4) −2(w2w3 + w1w4) w2

1 + w22 + w2

3 + w24 0

2(w2w3 + w1w4) 2(w1w3 − w2w4) 0 w21 + w2

2 + w23 + w2

4

and

(w21 + w2

2 + w23 + w2

4)2 − (2w1w3 − 2w2w4)

2 − (2w2w3 + 2w1w4)2 = 1.

HOLONOMY DISPLACEMENTS IN HOPF BUNDLES 737

Clearly, CH1 can be identified with the following

CH1 =

x 0 y z0 x −z yy −z x 0z y 0 x

: x2 − y2 − z2 = 1, x > 0

.

Therefore, the map

p : SU(1, 1) −→ CH1

defined by p(w) = ww has the following properties:

p(wv) = wp(v)w for all w, v ∈ SU(1, 1),

p(wv) = p(w) if and only if v ∈ S1.

This shows that the map p is, indeed, the orbit map of the principal bundleS1 → SU(1, 1) → CH1. The Lie group SU(1, 1) will have a left-invariantRiemannian metric given by the following orthonormal basis on the Lie algebrasu(1, 1)

e1=

0 0 1 00 0 0 11 0 0 00 1 0 0

, e2=

0 0 0 10 0 −1 00 −1 0 01 0 0 0

, e3=

0 1 0 0−1 0 0 00 0 0 −10 0 1 0

.

Notice that e1 and e2 correspond to [ 01 ] and [ 0i ] in gl(2,C) and [e1, e2] =−2e3. Consider the subset of SU(1, 1):

T =

[coshx (sinhx)e−iy

(sinhx)eiy coshx

]: x ≥ 0, 0 ≤ y ≤ 2π

=

coshx 0 (sinhx)(cos y) (sinhx)(sin y)0 coshx −(sinhx)(sin y) (sinhx)(cos y)

(sinhx)(cos y) −(sinhx)(sin y) coshx 0(sinhx)(sin y) (sinhx)(cos y) 0 coshx

which is the exponential image of

m =

[0 ξt

ξ 0

]: ξ ∈ C

.

Of course, SU(1, 1) is topologically a product S1×CH1. The map p restrictedto T is just the squaring map; that is,

p(w) = w2, w ∈ T.

Theorem 3.1. Let S1 → SU(1, 1) → CH1 be the natural fibration. Let γ bea piecewise smooth, simple closed curve on CH1. Then the holonomy displace-ment along γ is given by

V (γ) = e12A(γ)i ∈ S1,

where A(γ) is the area of the region on CH1 enclosed by γ.

738 YOUNGGI CHOI AND KYUNG BAI LEE

Proof. Let γ(t) be a closed loop on CH1 with γ(0) = p(I4). Therefore,

γ(t) =

cosh 2x(t) 0 sinh 2x(t) cos y(t) sinh 2x(t) sin y(t)

0 cosh 2x(t) − sinh 2x(t) sin y(t) sinh 2x(t) cos y(t)sinh 2x(t) cos y(t) − sinh 2x(t) sin y(t) cosh 2x(t) 0sinh 2x(t) sin y(t) sinh 2x(t) cos y(t) 0 cosh 2x(t)

.Let

γ(t) =

coshx(t) 0 sinhx(t) cos y(t) sinhx(t) sin y(t)

0 coshx(t) − sinhx(t) sin y(t) sinhx(t) cos y(t)sinhx(t) cos y(t) − sinhx(t) sin y(t) coshx(t) 0sinhx(t) sin y(t) sinhx(t) cos y(t) 0 coshx(t)

with x(t) ≥ 0 so that p(γ(t)) = γ(t) (γ is a lift of γ), and let

ω(t) =

cos z(t) − sin z(t) 0 0sin z(t) cos z(t) 0 0

0 0 cos z(t) sin z(t)0 0 − sin z(t) cos z(t)

.

Put η(t) = γ(t)·ω(t). We still have p(η(t)) = γ(t), and η is another lift of γ. Wewish η to be the horizontal lift of γ. That is, we want η′(t) to be orthogonal tothe fiber at η(t). The condition is that ⟨η′(t), (ℓη(t))∗(e3)⟩ = 0, or equivalently,⟨(ℓη(t)−1)∗η

′(t), e3⟩ = 0. That is,

η(t)−1 · η′(t) = α1e1 + α2e2

for some α1, α2 ∈ R. From this, we get the following equation:

(3–1) z′(t) = sinh2 x(t)y′(t).

By virtue of Lemma 2.1, it will be enough to prove the statement for aparticular type of curves as follows: Suppose we are given a rectangular regionin the xy-plane

p ≤ x ≤ p+ a, q ≤ y ≤ q + b.

Consider the image R of this rectangle in CH1 by the map

(x, y) 7→ r(x, y) = (cosh 2x, (sinh 2x)(cos y), (sinh 2x)(sin y)),

CH1 with the (+−−) metric. The area of R can be calculated as follows:

rx × ry = ((2 cosh 2x)(sinh 2x),−(2 sinh2 2x)(cos y),−(2 sinh2 2x)(sin y)).Now

||rx × ry|| = 2| sinh 2x|, (+−−)-norm= 2 sinh 2x (because x ≥ 0).

Thus, the area is∫ q+b

q

∫ p+a

p

2 sinh 2x dxdy =[[2 sinh2 x

]p+a

p

]q+b

q= 2b(sinh2(p+ a)− sinh2(p)).

On the other hand, the change of z(t) along the boundary of this regioncan be calculated using condition (3–1). Label the four vertices by A(p, q),

HOLONOMY DISPLACEMENTS IN HOPF BUNDLES 739

B(p + a, q), C(p + a, q + b), and D(p, q + b). AB can be parametrized byx(t) = p + at, y(t) = q, t ∈ [0, 1] so that y′(t) = 0. For BC, x(t) = p + a,y(t) = q + bt, t ∈ [0, 1]. Then

z(1)− z(0) =

∫ 1

0

z′(t)dt =

∫sinh2(p+ a)b dt = b · sinh2(p+ a).

Similarly, z(t) does not change along CD, but onDA, x(t) = p, y(t) = q+b−bt,t ∈ [0, 1]. So

z(1)− z(0) =

∫ 1

0

z′(t)dt =

∫sinh2(p)(−b)dt = −b · sinh2(p).

Thus the total vertical change of z-values, z(1) − z(0), along the perimeter ofthis rectangle is b · (sinh2(p+ a)− sinh2(p)) which is 1/2 times the area.

Now we turn to the general case

S1 −→ S2n,1 p−→ CHn.

We are viewing S2n,1 ∼= U(1, n)/U(n), and CHn ∼= U(1, n)/(U(1)×U(n)). TheLie algebra of U(1, n) is u(1, n), and has the following canonical decomposition:g = h+m, where

h = u(1) + u(n) =

[λ 00 B

]: λ+ λ = 0, B ∈ u(n)

and

m =

[0 ξt

ξ 0

]: ξ ∈ Cn

.

Lemma 3.2. A 2-dimensional subspace m′ of m ⊂ u(1, n) gives rise to a com-plete totally geodesic submanifold of CHn if and only if either

(1) m′ is J-invariant (i.e., has a complex structure), or(2) m′ has tangent vectors v and w such that vw − vw = 0.

Furthermore, for each of these cases, the pullback of the bundle S1 → S2n,1 →CHn by the inclusion is isomorphic to the standard bundle S1 → SU(1, 1) →CH1 for (1), or the product bundle S1 × CH1 for (2), respectively.

Proof. With the notation m as above, let v and w be elements of m whose ξ’sare given by

x1 + iy1x2 + iy2· · ·

xn + iyn

for v and

a1 + ib1a2 + ib2· · ·

an + ibn

for w.

Then by Proposition 2.2, m′ is a totally geodesic sub-manifold if and only if[[m′,m′],m′] ⊂ m′. Some calculations show the following equality

[[v,w],v] =n∑

k=1

(xkak + ykbk)v −n∑

k=1

(x2k + y2k)w − 3

n∑k=1

(xkbk − ykak)(iv)

740 YOUNGGI CHOI AND KYUNG BAI LEE

holds. Therefore, [[v,w],v] = pv + qw for some real p and q if and only ifiv = pv + qw has solution for some real p and q.

Suppose m′ has a complex structure. Then we can take v and w in m′

so that iv = w (so ak = −yk and bk = xk for all k = 1, 2, . . . , n). Thus,[[v,w],v] = pv+qw has a solution for p and q. Suppose vw−vw = 2Im(vw) =2∑n

k=1(xkbk− ykak) = 0. Then clearly [[v,w],v] = pv+ qw has a solution forp and q.

Conversely, suppose [[v,w],v] = pv + qw has a solution for p and q. Theniv = pv + qw must have a real solution for p and q. Suppose

∑nk=1(xkbk −

ykak) = 0. Then, at least one of the summands is non-zero, say x1b1−y1a1 = 0.This means that we can find a new basis for spanv,w with

x1 = 1, y1 = 0; a1 = 0, b1 = 1.

Then the equation pv + qw = iv is quickly reduced to p = 0 and q = 1 (fromk = 1), and hence we obtain xk = bk, yk = −ak for all k = 2, . . . , n. Thisshows w = iv, and the space spanned by v and w has a complex structure.

For the second part of the statement, it is enough to observe that

[v,w] =

[λ 00 0

],

where λ = vw − vw. If λ = 0, then the distribution m′ is integrable, and thebundle is trivial.

By combining Theorem 3.1 and Lemma 3.2, we have now:

Theorem 3.3. Let S1 → S2n,1 → CHn be the natural fibration. Let S be acomplete totally geodesic 2-dimensional surface in CHn, and ξS be the pullbackbundle over S. Let γ be a piecewise smooth, simple closed curve on S. Thenthe holonomy displacement along γ is given by

V (γ) = e12A(γ)i or e0i ∈ S1,

where A(γ) is the area of the region on the surface S surrounded by γ, dependingon whether S is a complex submanifold or not.

Since the length of η is half of the length of γ, we have:

Corollary 3.4. Suppose γ is a piecewise smooth, simple closed curve parametr-ized by arc length. Then the Hopf torus in S2n,1 over γ is isometric to the torusgenerated by the lattice (2π, 0), (A(γ)/2 , L(γ)/2 ) in R2, where L(γ) is thelength of γ.

4. The complex Heisenberg group H2n+1

We consider H2n+1. This is R× Cn with group operation given by

(s, z)(t, z′) = (s+ t+ 2 Imzz′, z+ z′),

where Imzz′ is the imaginary part of the complex number z1z′1 + z2z

′2 +

· · · + znz′n for z = (z1, z2, . . . , zn), z

′ = (z′1, z′2, . . . , z

′n) ∈ Cn. This is a 2-step

HOLONOMY DISPLACEMENTS IN HOPF BUNDLES 741

nilpotent Lie group with center Z(H2n+1) = R. In the case of n = 1, H3 isisomorphic to the ordinary 3-dimensional Heisenberg group by

(4z − 2xy, x+ iy)←→

1 x z0 1 y0 0 1

.

For the sake of computations, we use the following affine representation ofH2n+1 into Aff(2n+ 1) ⊂ GL(2n+ 2):

s,

x1 + iy1x2 + iy2

· · ·· · ·

xn + iyn

−→

1 −2y1 2x1 · · · −2yn 2xn s0 1 0 · · · 0 0 x1

0 0 1 · · · 0 0 y1· · · · · · · · ·· · · · · · · · ·0 0 0 · · · 1 0 xn

0 0 0 · · · 0 1 yn0 0 0 · · · 0 0 1

.

The Lie algebra has a following orthonormal basis

e1=

0 0 2 · · · 0 0 00 0 0 · · · 0 0 10 0 0 · · · 0 0 0· · · · · · · · ·· · · · · · · · ·0 0 0 · · · 0 0 00 0 0 · · · 0 0 00 0 0 · · · 0 0 0

, e2=

0 −2 0 · · · 0 0 00 0 0 · · · 0 0 00 0 0 · · · 0 0 1· · · · · · · · ·· · · · · · · · ·0 0 0 · · · 0 0 00 0 0 · · · 0 0 00 0 0 · · · 0 0 0

,

......

e2n=

0 0 0 · · · −2 0 00 0 0 · · · 0 0 00 0 0 · · · 0 0 0· · · · · · · · ·· · · · · · · · ·0 0 0 · · · 0 0 00 0 0 · · · 0 0 10 0 0 · · · 0 0 0

, e2n+1=

0 0 0 · · · 0 0 10 0 0 · · · 0 0 00 0 0 · · · 0 0 0· · · · · · · · ·· · · · · · · · ·0 0 0 · · · 0 0 00 0 0 · · · 0 0 00 0 0 · · · 0 0 0

which defines a left-invariant Riemannian metric on H2n+1. The short exactsequence of groups

1 −→ R −→ H2n+1 p−→ Cn −→ 1

is a fiber bundle, which is topologically trivial. The left invariant metric natu-rally induces a connection on this principal R-bundle. There is a unique metricon Cn (standard Euclidean metric) which makes the projection map p a Rie-mannian submersion.

742 YOUNGGI CHOI AND KYUNG BAI LEE

Theorem 4.1. Let 1 → R → H2n+1 p→ Cn → 1 be the central short exactsequence of the complex Heisenberg group. Let S be a complete totally geodesicplane in Cn, and ξS be the pullback bundle over S. Let γ be a piecewise smooth,simple closed curve on S. Then

V (γ) = e(ξS) ·A(γ),where A(γ) is the area of the region on the surface S surrounded by γ, andthe number e(ξS) is determined by the equality [v,w] = e(ξS)e2n+1 for anorthonormal basis v,w for the tangent space of S.

Proof. Every complete totally geodesic submanifold of Cn is an R-linear sub-space of Cn. Therefore S = spanv,w for some orthonormal basis v,w wherev =

∑nj=1(aj + ibj) and w =

∑nj=1(cj + idj) ∈ Cn. Then γ is of the form

γ(t) = x(t)v + y(t)w ∈ S ⊂ Cn,

where x(t) and y(t) are scalars. We want to find a curve z(t) in R so thatη(t) = (z(t), γ(t)) is orthogonal to the fiber for every t. In other words,

⟨η′(t), (ℓη(t))∗e2n+1⟩ = 0,

where ℓ is the left translation. This is equivalent to ⟨(ℓ−1η(t))∗η

′(t), e2n+1⟩ = 0.

Note that η(t)−1 = (−z(t) + 2 Imγ(t)γ(t),−γ(t)). Using the affine represen-tation, (ℓη(t)−1)∗η

′(t) is

0 −2(x′(t)b1 + y′(t)d1) · · · z′(t)− 2(x(t)y′(t)− x′(t)y(t)) Imvw0 0 · · · x′(t)a1 + y′(t)c10 0 · · · x′(t)b1 + y′(t)d1· · · · · ·· · · · · ·0 0 · · · x′(t)an + y′(t)cn0 0 · · · x′(t)bn + y′(t)dn0 0 · · · 0

,

where Imvw =∑n

j=1(ajdj − cjbj). Note that Imvw = Imv′w′ for anyorthonormal basis v′,w′. The equation ⟨(ℓη(t)−1)∗η

′(t), e2n+1⟩ = 0 gives riseto

(4–1) z′(t)− 2(x(t)y′(t)− x′(t)y(t)) Imvw = 0.

Suppose we are given a rectangular region on xy-plane

p ≤ x ≤ p+ a, q ≤ y ≤ q + b.

Consider the image R of this rectangle in S ⊂ Cn by the map

(x, y) 7→ xv + yw.

Then R is a rectangle with vertices pv+qw, (p+a)v+qw, (p+a)v+(q+b)w,pv + (q + b)w. Let γ(t) be the piecewise linear boundary curve. It can berepresented by ((p+4at)v, qw) for 0 ≤ t ≤ 1/4, ((p+ a)v, (q+ b(4t− 1))w) for

HOLONOMY DISPLACEMENTS IN HOPF BUNDLES 743

1/4 ≤ t ≤ 1/2, ((p+a(3−4t))v, (q+b)w) for 1/2 ≤ t ≤ 3/4, (pv, (q+b(4−4t))w)for 3/4 ≤ t ≤ 1.Then, from the equation (4–1),

z(1)− z(0) = 2

∫ 1

0

(x(t)y′(t)− x′(t)y(t)) Imvwdt = 4ab Imvw.

On the other hand,

[v,w] = 4

0 0 0 · · · 0 0 Imvw0 0 0 · · · 0 0 00 0 0 · · · 0 0 0· · · · · · · · ·· · · · · · · · ·0 0 0 · · · 0 0 00 0 0 · · · 0 0 00 0 0 · · · 0 0 0

.

This means that [v,w] = 4 Imvwe2n+1 = e(ξS) e2n+1 so that V (γ) = e(ξS) ·A(γ) with e(ξS) = 4 Imvw.

Having shown the statement for rectangular regions, now we apply Lemma2.1 to conclude that the same formula holds for any piecewise smooth, simpleclosed curve. Corollary 4.2. Suppose γ is a piecewise smooth, simple closed curve parametr-ized by arc-length. Then the Hopf cylinder in H2n+1 over γ is isometric to thecylinder generated by the translation (x, y) 7→ (x + e(ξS)A(γ), y + L(γ) ) onR2, where L(γ) is the length of γ.

References

[1] S. Kobayashi and K. Nomizu, Foundations of differential geometry, Vol. II, Reprint ofthe 1969 original. Wiley Classics Library, A Wiley-Interscience Publication, John Wiley

& Sons, Inc., New York, 1996.[2] U. Pinkall, Hopf tori in S3, Invent. Math. 81 (1985), no. 2, 379–386.[3] J. Wolf, Spaces of Constant Curvature, 5th ed., Publish or Perish, Wilmington, 1984.

Younggi Choi

Department of Mathematics EducationSeoul National UniversitySeoul 151-748, KoreaE-mail address: yochoi@snu.ac.kr

Kyung Bai LeeDepartment of MathematicsUniversity of Oklahoma

Norman, OK 73019, USAE-mail address: kblee@math.ou.edu